ACSBR 2024 4E5N Prelim MATH P1 solutions

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) Qn Steps/Answer Remarks 1 81.0 0.902 3 2 86.0 441 399 2 5 1 )5( 7 2 −−−= xx x Also accept 2 71 (5 ) 5 x xx +−− 2 7 ( 5) ( 5) xx x −−= − 2 75 (5 ) xx x +−= − 2 65 ( 5) x x += − 2 65 (5 ) x x += − 3 96(254.9 10 ) (2.45 10 )   Accept 96(254 10 ) (2.45 10 )   or higher accuracy = 51.04 10 4 25 20 2 cc d 2 5 2 20 cd c= 8 d c= 5 22 2 9 100%kx kx kx −  = 800% 6 (3-2) units -> $20 Total 9 units -> $180 7 (a) x328 −− and 12 3 7 2xx−  − 10 3 x− − and 152 2 x− 3 132 − x cannot accept 3.33 (b) -1, 0, 1, 2, 3 8 (a) (bi) 7  B (bii)  3,7  A

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 9 Construction arcs are to be clearly seen. Correct angle bisector Correct perpendicular bisector Correct region shaded 10 (2 1)( 2) 0pp+ − = Accept (2 1)(3 6) 0pp+ − = 1 ,22pp=− = 11 Cost price of watch for Jimmy = 210100 80  168$= Profit price = 168$100 120 60.201$= Marked price = 60.201$90 100 = $224 12 (a) Total time taken = 15 5 5++ = 25 min Yes, he will achieve his target as he will complete by 09 15. (b) 2.3 1000 25  = 92 m/min 13 (a) 3200 1.49() 1V = V = 60g (b) Different vertical scales/intervals are used. 14 (a) 221260 2 3 5 7=    (b) 84 and 180 (c) 3, 2mn==

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 15 (a) 2 2 1 12 1 15 4 16 xx x ++ = + + (b) 16 (a) V =           130 100 120 410 380 320 (b) C =       5.0 2 (c) P =           885 810 700 (d) A represents the average ERP charges collected across the three days. 17 (a) 1 202 121 (2 ) 402 base width base width         or 2 20 1240   = 3 min (b)

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 18 4979.97.0tan 8 == radQP Area of triangle OPQ = 992.37)4979.9)(8(2 1 = Area of sector = ( ) 21 8 0.7 27.86522  −= Area of shaded region = 10.1 cm2 19 (a) 11(6) (18 6) (23 18) 38522 v v v+ − + − = 22 m/sv= (b) 30 0 25 0 35 45 35T −− =−− or 10 22 = 30(10) 25( 35)T=− 47sT = or 45+2=47 20 (a) angle PBC = angle QBR (common angle) angle BQR = angle BPC (corr. angles, PC//QR) Triangle PCB and triangle QRB are similar (AA test) (b) Triangles ABC and QRC or Triangles ABQ and CPQ (c) 3 7 CR CB = 34 7 QR BR PC BC PC = = 5.25PC=

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 21 (a) 1 6 2 4 4a b −   1 4 2 64 b a = 2 32 b a= (b) 2 4 2 4 8 k k= 4 3 4 2 2 2 k k= 34 4kk=+ 1 4k =− 22 (a) 2 ( 1) 10 4( 1) 10 4 4 nT n n n n n n = + + − − = + + − + 2 3 14nn= − + (b) 236450 =T (c) 14)3(1432 +−=+− nnnn When n is even n(n-3) is (even x odd) = even. When n is odd, n(n-3) is (odd x even) = even. Adding to 14 which is also even, 1432 +−= nnTn will always be even for all terms. 23 (a) ( )5 2 180 1085EDC BAE − = = =  180 108 362AEB − = =  108 36 72BED = − =  72 108 180BED EDC + = + =  By the converse of interior angles, BE is parallel to CD (b) 180 36 144EBX = − =  Or equivalent methods, with correct reasoning. (eg BE//CX and a pair of opposite equal angles) 108 36 36 36BEC = − − =  As 180EBX BEC + =  , by the converse/property of interior angles, EC//BX. BECX is a rhombus as BE//CX (or DX) and EC//BX and adjacent sides BE=EC.

Mathematics Paper 1 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 24 (ai) 5 (aii) Gradient of PQ = 3 4 − 32 (8)4 c−− = + c = 4 Equation is 3 44yx −=+ (b) 32 4 4 194 xx− + − = Or elimination method correctly follow-thru from a(ii) 2, 5.5xy=− = ( 2,5.5)− 25 (a) (b) 12 11 11 25 24 50     =         (c) 13 12 225 24         = 13 25   (d) 12 11 10 13 12 111 25 24 23 25 24 23        −−               or P(BBG)+P(BGB)+P(GBB)+P(GGB)+P(GBG)+P(BGG) = 12 11 13 13 12 123325 24 23 25 24 23         +                = 0.78 or 39 50  